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Math Help - Second order DE help

  1. #1
    Member Jones's Avatar
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    Second order DE help

    Hi,

    I have the equation:
    \begin{cases}y^{\prime\prime} + 3y^{\prime} + 2y = sin~x \\<br />
y(0) = 1 \\<br />
y^{\prime}(0) = 1 \end{cases}

    The auxiliary equation t^2 + 3t + 2 = 0 has the solutions
    x_{1}=-1~~ x_{2}=-2

    So the general solution would look like this
    Ae^{-t}+Be^{-2t}

    Now, let \begin{cases}y = A~Cos(t)+B~Sin(t) \\<br />
y^{\prime} = -A~Sin(t)+B~Cos(t) \\<br />
y^{\prime\prime} = -A~Cos(t)-B~Sin(t) \end{cases}

    gives -A~Cos(t)-B~Sin(t) + 3\cdot(-A~Sin(t)+B~Cos(t)) + 2\cdot(A~Cos(t)+B~Sin(t))

    Can be simplified down to 3A~Cos(t)-3B~Sin(t) + ASin(t) +B~Cos(t)

    This is where i am stuck, how do you combine this with the general solution?

    //Thanks
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  2. #2
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    Well, you know that 3A+B=0 since there are no cos x terms, and -3B+A=0 since there is exactly one sin x term.

    As a sidebar, it's easier to answer things like this clearly when you use standard notations. Your initial equation indicates x as the independent variable, and \lambda is what is typically used for calculating the characteristic equations.
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  3. #3
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    That doesn´t look right

    If 3A + B = 0 (1) -> = B + 3A = 0
    and  B - 3A = 0 (2) -> = B - 3A = 0 then A must be 0

    From (1) -> B = -3A
    Substituting into (2) -> -3A - 3A = 0 -> -6A = 0 then A = 0

    And from (2) -> B = 3A substituting into (1) ->  B + B = 0 then B must be 0

    Since you have Ae^-t + Be^-2t you would get y;y`= 0
    Hope it was somewhat helpful
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  4. #4
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    Accually you can

    Was the textbook answer -0.5e^(-t) + -0.5e^(-2t) -0.5cosx -0.5sinx???
    Last edited by Henryt999; January 11th 2010 at 03:25 AM.
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  5. #5
    Member Jones's Avatar
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    Quote Originally Posted by Henryt999 View Post
    Was the textbook answer -0.5e^(-t) + -0.5e^(-2t) -0.5cosx -0.5sinx???
    I don't know, i don't have the correct answer...

    Anyway,

    we have \begin{cases}3A + B = 0 \\<br />
-3B + A = 1\end{cases}

    A = 0.1
    B = -0.3

    So y_p = 0.1Sin(t)-0.3Cos(t)
    But how do i pair this up with the general solution?
    Last edited by Jones; January 11th 2010 at 05:05 AM.
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  6. #6
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    Quote Originally Posted by Jones View Post
    I don't know, i don't have the correct answer...

    Anyway,

    we have \begin{cases}3A + B = 0 \\<br />
-3B + A = 1\end{cases}

    A = 0.1
    B = -0.3

    So y_p = 0.1Sin(t)-0.3Cos(t)
    But how do i pair this up with the general solution?
    Just add it to the general solution of the homogeneous equation!

    If L(y) is a linear differential operator, then L(y+ y2)= L(y1)+ L(y2). If yh satifies L(yh)= 0 and yp satisfies L(yp)= f(x), then L(yh+ yp)= L(yh)+ L(yp)= 0+ f(x)= f(x). Further, if y is any solution to L(y)= f(x), then L(y- yp)= L(y)- L(yp)= f(x)- f(x)= 0 so that y- yp= yh, a solution to the homogeneous equation. The general solution to the entire equation is the the general solution to the homogeneous equation plus any solution to the entire equation. That's the whole point of this method!
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